The linear mass density of a rod of length L | vedclass

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(C) Let the rod be placed along the $x$-axis with one end at the origin. Since the rod lies on the $x$-axis,the $y$ and $z$ coordinates of the center

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$\frac{L(3A + 2BL)}{3(2A + BL)}$

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(C) Let the rod be placed along the $x$-axis with one end at the origin. Since the rod lies on the $x$-axis,the $y$ and $z$ coordinates of the center of mass are zero,i.e.,$y_{CM} = 0$ and $z_{CM} = 0$.Consider a small element of length $dx$ at a distance $x$ from the origin. The mass of this small element is $dm = \lambda dx = (A + Bx)dx$.The $x$-coordinate of the center of mass is given by:$x_{CM} = \frac{\int_{0}^{L} x dm}{\int_{0}^{L} dm} = \frac{\int_{0}^{L} x(A + Bx) dx}{\int_{0}^{L} (A + Bx) dx}$Evaluating the integrals:Numerator: $\int_{0}^{L} (Ax + Bx^2) dx = [\frac{Ax^2}{2} + \frac{Bx^3}{3}]_{0}^{L} = \frac{AL^2}{2} + \frac{BL^3}{3} = \frac{3AL^2 + 2BL^3}{6} = \frac{L^2(3A + 2BL)}{6}$Denominator: $\int_{0}^{L} (A + Bx) dx = [Ax + \frac{Bx^2}{2}]_{0}^{L} = AL + \frac{BL^2}{2} = \frac{2AL + BL^2}{2} = \frac{L(2A + BL)}{2}$Therefore,$x_{CM} = \frac{L^2(3A + 2BL)}{6} 	imes \frac{2}{L(2A + BL)} = \frac{L(3A + 2BL)}{3(2A + BL)}$.

The linear mass density of a rod of length $L$ varies as $\lambda = A + Bx$. Find the center of mass.

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